Potential dividersEdexcel International A Level Physics: Revision notes

Section 1

The potential divider principle

A potential divider is two (or more) resistors in series across a supply. The same current $I$ flows through each, so the supply p.d. is shared in proportion to the resistances. For resistors $R_1$ and $R_2$ across a supply of p.d. $V_{in}$ with the output taken across $R_2$ :

$V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}$

This follows from $I = V_{in}/(R_1 + R_2)$ and $V_{out} = IR_2$ . The larger resistor takes the larger share of the p.d. The output voltmeter is assumed to have a very high resistance, so it draws negligible current.

Key termspotential divideroutput p.d.

Common mistake

The output is across the resistor in the numerator. Taking it across R₁ instead gives V_in × R₁/(R₁ + R₂).

Section 2

Calculating p.d.s and resistances

Calculations use either the ratio $V_{out} : V_{in} = R_2 : (R_1 + R_2)$ or the current. To find a missing resistor, work out the current from the known resistor ( $I = V/R$ ), find the p.d. across the unknown ( $V_{in} - V_{out}$ ) and divide: $R = V/I$ .

For example, 9.0 V across 2.0 kΩ and 4.0 kΩ: $V_{out} = 9.0 \times 4.0/6.0 = 6.0$ V. The power in each resistor is $P = VI$ , and the total equals the power supplied by the source.

Key termsseries circuit

Exam tip

Check your answer: the p.d.s across the two resistors must add up to the supply p.d.

Section 3

Variable resistance and sliding contacts

If one resistor is variable, the output changes. A variable resistor used as a potential divider with a sliding contact gives an output from 0 V to the supply p.d. as the contact moves. If the variable resistor is the lower one (output across it), increasing its resistance increases the output; if it is the upper one, increasing its resistance decreases the output.

Key termsvariable resistor

Section 4

Thermistors and LDRs in dividers

An NTC thermistor has a resistance that decreases as temperature rises. An LDR has a resistance that decreases as light intensity rises.

In a divider with the sensor as the upper resistor and a fixed resistor below it (output across the fixed resistor), when the sensor's resistance falls (hotter or brighter), the fixed resistor takes a larger share and the output increases. With the output taken across the sensor, the output decreases as it gets hotter or brighter. Swapping the two components reverses the effect, which lets the designer choose a circuit that switches on when it gets hotter or darker.

Key termsNTC thermistorLDR

Common mistake

State which component the output is across before saying whether the output rises or falls.

Section 5

Worked example

A 6.0 V supply is connected across an LDR and a 10 kΩ resistor in series. The output is across the 10 kΩ resistor.

Bright light, LDR 2.0 kΩ: $V_{out} = 6.0 \times 10/12 = 5.0$ V
Dim light, LDR 40 kΩ: $V_{out} = 6.0 \times 10/50 = 1.2$ V

The output falls as the light intensity falls.

Must Know

V_out = V_in × R₂/(R₁ + R₂); same current, p.d. shared in proportion to R
Find unknown R from I = V/R of the known resistor
NTC thermistor and LDR: resistance falls as temperature or light rises
Output across the sensor falls as it gets hotter or brighter; across the fixed resistor it rises
The voltmeter is assumed to have very high resistance

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Potential dividers

A 9.0 V supply of negligible internal resistance is connected across a 2.0 kΩ resistor in series with a 4.0 kΩ resistor. A voltmeter of very high resistance measures the output p.d. across the 4.0 kΩ resistor.
Calculate the current in the resistors and the power dissipated in the 4.0 kΩ resistor.2 marks
A light-meter circuit has a 6.0 V supply of negligible internal resistance, a light-dependent resistor (LDR) and a 10 kΩ fixed resistor, all in series. The output p.d. is measured across the fixed resistor with a voltmeter of very high resistance. In bright light the LDR has a resistance of 2.0 kΩ; in dim light its resistance is 40 kΩ.
Calculate the p.d. across the LDR in dim light.2 marks
A temperature-sensing circuit has an NTC thermistor in series with a 4.0 kΩ fixed resistor across a 5.0 V supply of negligible internal resistance. The output p.d. is taken across the fixed resistor. At 20 °C the resistance of the thermistor is 12 kΩ and at 60 °C it is 1.0 kΩ.
Calculate the output p.d. of the circuit at 20 °C.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).